straight lines

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Straight Lines

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Page 1: straight lines

Straight Lines

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•Basics of Coordinate Geometry

1. Coordinate axes2. Coordinate plan3. Plotting of points in plane4. Distance between two points5. Section Formula

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•Slope of a Line1. If θ is the inclination of a line L,then θ

is called the slope or gradient of the line L.

2. The slope of a line is denoted by m. Thus,m=tan θ,θ≠90°

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θ180°-θ

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•Slope of a line when coordinates of any two points on the line.

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Let P (X₁,Y₁)and Q(X₂,Y₂) be two points on non-vertical line l whose inclination is θ.Obviously, X₁≠X₂. otherwise the line will become perpendicular to X-axis and its slope will not be defined. The inclination of the line l may be acute or obtuse. Draw QR perpendicular to x-axis and PM perpendicular to RQ. Let us take two case.

l Q(X₂,Y₂)

P(X₁,Y₁)

θ

θM

ROX

Y

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Case:- 1 When angle θ is acute:

In fig, MPQ=θ.Therefore,slope of line l=m=tanθ. .......(1)

But in ∆MPQ,we have tanθ=MQ/MP = Y₂-Y₁/X₂-X₁ ..........(2)From (1) &(2),we havem= Y₂-Y₁/X₂-X₁

˂

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Case :- 2Y

XO

M

Q(X₂,Y₂)

P(X₁,Y₁)θ

180°-θ

R

l

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When angle θ is obtuse: In fig, we have m=tanθ MPQ=180°-θ. Therefore,θ=180°- MPQNow, slope of line =tan(180°- MPQ) =-tan MPQ=- MQ/MP = - (Y₂-Y₁)/(X₁-X₂) = Y₂-Y₁/X₁-X₂Consequently, we see that in the both cases the

slope m of line through the points (X₁,Y₁)&(X₂,Y₂) is given by m= Y₂-Y₁/X₁-X₂

˂

˂˂

˂

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Condition for parallelism and perpendicularity of line in terms of their slopes•If the line l is parallel to L then their inclinations are equal,i.e.., =β,and hence, tan=tanβ Therefore m₁=m₂, i.e.,their slopes are equal.Conversly,if the slope of two lines l₁ & l₂ is same,i.e., m₁=m₂Then tan=tanβBy property of tangent function (between 0° & 180° ,=β.Therefore,the line are parallel if and only if their slopes are equal.

Y

XO

l₁l₂

β

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If the lines L & l are perpendicular then β= + 90°.Therefore,tan β =tan(+ 90°) = -cot = - 1/tan i.e., m₂=- 1/m₁ or m₁.m₂=-1Conversely, if m₁.m₂ =-1, i.e,tantanβ=-1 Then tanβ=cotβ=tan(β+90°) or tan(β-90°)Therefore &β differ by 90°.Hence, two non vertical lines are perpendicular to each other if and only if their slopes are negativereciprocals of each other.

Y

XO

β

l₁l₂

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•Angle between two linesLet L₁ & L₂ be two non vertical lines with slopes m₁ & m₂ respectively. if &β are the inclination of lines L₁&L₂ respectively. Then m₁=tan₁ & m₂=tan₂. Let θ & be adjacent angles between the line L₁ & L₂. Then θ=₂-₁ & ₁,₂≠90°.Therefore tanθ=tan(₂-₁) =tan₂-tan₁/1+tan₁tan₂ =m₂-m₁/1+m₁m₂And =180°-θ tan=tan(180°-θ) = - (m₂-m₁)/1+m₁m₂Now there arise two cases.

Y

XO

₁ ₂

θ

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case:-1If tanθ is positive, then tan will be negative, which means that θ will be acute and will be obtuse.

case:-2If tanθ is negative,then tan will be positive, which means that θ will be obtuse and will be acute.Thus, for solve problem we use tanθ=│m₂-m₁/1+m₁m₂│.

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•Collinearityof three point We know that slopes of two parallel lines are equal. If two lines having the same slope pass through a common point, then two lines will coincide.hence,if A,B &C are three points in the XY plane, then they will lie on a line ,i.e., three points are collinear if and only slope of AB= slope of BC

A

B

C

X

Y

O

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•Various Forms of the equation of a line

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•Horizontal and vertical linesIf a horizontal line L is distance a from the X-axis then ordinate of every point lying on the line is either a or –a. Therfore,equation of line L is either y=a or y=-a. sign will depend upon the position of the line according as the line is above or below the y-axis. Similarly, the equation of a vertical line at a distance b from the y-axis is either x=b or x= -b.

a

a Y=-a

Y=aL

L

b

X=-b X=b

-b

L L

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•Point-slope formSuppose that P₀(X₀,Y₀) is a fixed point on a non-vertical line L,whose slope is m.Let P(X,Y)be an arbitrary point on L.Then, by definition , the slope of L is given by m=Y-Y₀/X-X₀ Y-Y₀=m(X-X₀) …(1) Since the point P₀(X₀,Y₀) along with all points (X,Y) on L satisfies (1) and no other point in plane satisfies (1).Equation (1) is needed the equation for the given line L. Thus, the point (X,Y) lies on the line with slope m through the fixed point(X₀,Y₀), if and only if, its coordinates satisfies the equation Y-Y₀=m(X-X₀)

P₀(X₀,Y₀)

P(X,Y)

Slope m

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•Two-point formLet the line L passes through two given points P₁(X₁,Y₁) and P₂(X₂,Y₂). Let P(X,Y) be a general point on L.The three points P₁,P₂ & P are collinear , therefore, we have slope of P₁P=slope of P₁P₂ i.e., Y-Y₁/X-X₁=Y₂-Y₁/X₂=X₁ Thus , equation of the line passing through the points (X₁,Y₁) and (X₂,Y₂) is given by Y-Y₁/X-X₁=Y₂-Y₁/X₂-X₁

P₁(X₁,Y₁)

P(X,Y)

P₂(X₂,Y₂)

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•Slope-intercept form

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Case:- 1Suppose a line L with slope m cuts the y-axis at a distance c from the origin. The distance c is called the y-intercept of the line L. obvisouly,coordinate of the point where the line meet they-axis are (0,c).Thus , L has slope m and passes through a fixed point (0,c).Therefore , by point-slope form, the equation of L is y-c=m(x-0) or y=mx+cThus, the point (x,y) on the line with slope m and y-intercept c lies on the line if and only if y=mx+c Slop

e m (0,c)

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Case:-2•Suppose line L with slope m makes x-intercept d. Then equation of L is y=m(x-d)

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•Intercept - formSuppose a line L makes x-intercept a and y-intercept b on the axes.Obivously L meets x-axis at point (0,b) . By two form of the equation of the line we have y-0=b-0/0-a. (x-a)Thus, equation of the line making intercepts a and b on x-axis and y-axis , respectively, is x/a+ y/b=1

a

b

(0,b)

(a,0)

L

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•Normal form Suppose a non vertical line is known to us with following data:(1)Length of the e perpendicular from origin to the line.(2)Angle which normal makes with the positive direction of x-axis Let L be the line, whose perpendicular distance from origin O be OA=p and the angle between the positive to perpendicular x-axis and OA be XOA=ω. The possible position of the L in the Cartesian plan. Now , our purpose is to find slope of L and a point on it. Draw perpendicular AM on the x-axis in each case.

˂

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Figure:-1

A

M

ω

p

L

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Figure:-2

p

ω

A

M

L

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Figure:-3L ω

p

M

A

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Figure:-4

A

M

L

ω

p

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In each case, we have OM=pcosωand MA=psinω, so that the coordinates of the point A. Further , line L is perpendicular to OA.Therefore the slope of the line L=- 1/slope of OA =- 1/tanω = - cosω/sinω.Thus, the line L has slope -cosω/sinω and point A on it. Therefore , by point slope form , the equation of the line L is y- p sinω = - cos ω/ sinω(x-p cosω) x.cosω + y sinω=pHence, the equation of the line having normal makes distance p from origin and angle ω which the normal makes with the positive direction of x-axis is given by x.cosω + y.sinω=p

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•General Equation of a line

Ax+By+C=0 is general form equation of a line

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•Different forms of Ax+By+C=0

•If B≠0,then Ax+By+C=0 can be written as y=-A/B.x-C/B or y=mx+c ………(1) where m=-A/B and c=-C/B. we know that equation (1) is the slope intercept form of the equation of a line whose slope is -A/B and y-intercept is –C/B. If B=0,then x=-C/A, which is vertical line whose slope is undefined and x-intercept is –C/A.

(a) Slope intercept form

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•Intercept form

If C≠0,then Ax+By+C=0 can be written as x + y = 1 -C/A -C/B a=-C and b=-C A B we know that equation is the intercept form of the equation of a line whose x-intercept is –C/A and y-intercept is –C/B. If C=0,then Ax+By+C=0 can be written as Ax+By+C=0,which is a line passing through the origin and, therefore , has zero intercept on the axis.

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•Normal form Let x cosω+ y sinω=p be the normal form of the line represented by the equation Ax+By+C=0 or Ax+By=-C. Thus, both the equation are same and therefore, A = B =-C cosω sinω p which gives cosω=-Ap/C and sinω =-Bp/C. Now,sin² ω + cos² ω=(-Ap/C)² + (-BP/c)²=1 p²= C² or p=± C . A²+B² √A²+B²Therefore cosω=± A and √A²+B² Sinω=± B . √A²+B² Thus ,the normal form equation is xcosω+Ysinω=p

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•Distance of a point from a line The distance of a line is the length of the perpendicular drawn from the point to the line . let L:-Ax+By+C=0 be line , whose distance from the point P(x₁,y₁) is d. Draw a perpendicular PM from the point P to the L. If the line meets the x- and y-axes at the points Q and R, respectively. Then , coordinates of the points are Q(-C/A,0) and (0,-C/B).

R(0,-C/B)

P(x₁,y₁)

Q(-C/a,0)

d

M

Ax+By+C=0

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•Thus ,the area of the triangle PQR is given by area(∆PQR)=½PM.QR PM=2area (∆PQR) QR ………(1)also, area(∆PQR) =½│x₁(0+C/B)+(-C/A)(-C/B-y₁)+0(y₁-0)│ =½│x₁.C/B + y₁.C/A+C²/AB│ 2 area (∆PQR)=│C/AB│.│AX₁+By₁+C│, and QR= (0+C/A)²+ (C/B-0)² = │C/AB│.√A²+B²Substituting the values of area(∆PQR) and QR in (1) , we get PM=│Ax₁+By₁+C│/√ A²+B² d = │Ax₁+By₁+C│/√ A²+B².

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•Distance between two parallel linesWe know that slopes of two parallel lines are equal.Therefore , two parallel lines can be taken in the form y=mx+c₁ … (1)y=mx+c₂ ….(2)Line (1) will be intersect x-axis at point A(-c₁/m,0) .Distance between two lines is equal to the length of perpendicular from point A to line(2).Therefore , distance between the lines (1) and (2) is │(-m)(-c₁/m)+(-c₂)│/√1+m²d= │c₁-c₂│/ √1+m²Thus , the distance d between two parallel line y=mx+c₁ and y=mx+c₂ is given by d= │c₁-c₂│/ √1+m²If lines are given form , i.e. , Ax+By+C₁=0 & Ax+By+C₂=0 then d=│c₁-c₂│/√A²+B².

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Y=mx+c₁

Y=mx+c₂

d

A(-c₁/m,0)

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